Let $R$ be the region enclosed by the line $y=1$, the line $x=8$, and the curve $y=\text{log}_2(x)$. $y$ $x$ ${y=\text{log}_2(x)}$ ${y=1}$ $x=2}$ $(2,1)$ $(8,3)$ $(8,1)$ $ R$ A solid is generated by rotating $R$ about the line $x=2$. Which one of the definite integrals gives the volume of the solid? Choose 1 answer: Choose 1 answer: (Choice A) A $\pi\int_2^3 \left( -2^{2y}+4\cdot 2^y+32 \right) dy$ (Choice B) B $\pi\int_1^3 \left( -2^{2y}+4\cdot 2^y+32 \right) dy$ (Choice C) C $\pi\int_2^8 \left( -2^{2y}+4\cdot 2^y+32 \right) dy$ (Choice D) D $\pi\int_1^8 \left( -2^{2y}+4\cdot 2^y+32 \right) dy$
Let's imagine the solid is made out of many thin slices. Each slice is a cylinder with a hole in the middle, much like a washer. $y$ $x$ ${y=\text{log}_2(x)}$ ${y=1}$ $x=2}$ $(2,1)$ $(8,3)$ $(8,1)$ Let the thickness of each slice be $dy$, let the radius of the washer, as a function of $y$, be $r_1(y)$, and let the radius of the hole, as a function of $y$, be $r_2(y)$. Then, the volume of each slice is $\pi[(r_1(y))^2-(r_2(y))^2]\,dy$, and we can sum the volumes of infinitely many such slices with an infinitely small thickness using a definite integral: $\int_a^b \pi [(r_1(y))^2-(r_2(y))^2]\,dy$ This is called the washer method. What we now need is to figure out the expressions of $r_1(y)$ and $r_2(y)$, and the interval of integration. $r_1(y)$ is equal to the distance between the line $x=8$ and the line $x=2$. So, ${r_1(y)=6}$. $r_2(y)$ is equal to the distance between the curve $y=\text{log}_2(x)$ and the line $x=2$. To find it, we need to solve the equation for $x$ : $x=2^y$ So, ${r_2(y)=2^y-2}$. Now we can find an expression for the area of the washer's base: $\begin{aligned} &\phantom{=} \pi [({r_1(y)})^2-({r_2(y)})^2] \\\\ &= \pi [({6})^2-({2^y-2})^2] \\\\ &= \pi\left[ 36-(2^{2y}-4\cdot 2^y+4) \right] \\\\ &= \pi\left( -2^{2y}+4\cdot 2^y+32 \right) \end{aligned}$ The bottom endpoint of $R$ is at $y=1$ and the top endpoint is at $y=3$. So the interval of integration is $[1,3]$. Now we can express the definite integral in its entirety! $\begin{aligned} &\phantom{=}\int_1^3 \pi\left( -2^{2y}+4\cdot 2^y+32 \right) dy \\\\ &=\pi\int_1^3 \left( -2^{2y}+4\cdot 2^y+32 \right) dy \end{aligned}$